1. ## Parametric Arc Length

Alright I'm pretty lost here. I can usually deal with these types of problems fine, but with the given equations I don't seem to be getting anywhere.

Given two parametric functions:

x(t)=5e−05tsin(2t)
y(t)=5e−05tcos(2t)

0t6.

Find the integral.
Find the arc length, 0t6.
Find the arc length, [0).

Getting the derivative of each term and squaring it yields some nasty numbers and I'm running out of submissions.

The equation to go with that I've been using is:

2. Originally Posted by apples12
Alright I'm pretty lost here. I can usually deal with these types of problems fine, but with the given equations I don't seem to be getting anywhere.

Given two parametric functions:

x(t)=5e−05tsin(2t) are the bold expressions exponents?
y(t)=5e−05tcos(2t)

0t6.

Find the integral.
Find the arc length, 0t6.
Find the arc length, [0).

Getting the derivative of each term and squaring it yields some nasty numbers and I'm running out of submissions.

The equation to go with that I've been using is:

...

3. sorry my formatting got messed up in the posting

x(t)=sin(2t)5e^(-0.5t)
y(t)=cos(2t)5e^(-0.5t)

4. $x(t) = 5 e^{t/2} \sin 2t$

$y(t) = 5 e^{t/2} \cos 2t$

$x' = 5(1/2 e^{t/2} \sin 2t + 2e^{t/2} \cos 2t)$

$x' = 5e^{t/2} (1/2 \sin 2t + 2\cos 2t )$

in the same way

$y' = 5e^{t/2} (1/2 \cos 2t - 2\sin 2t )$

ok

$x'^2 + y'^2 = 25e^t ( 1/4 \sin ^2 2t + 2\sin 2t \cos 2t + 4 \cos ^2 2t +$ $1/4 \cos ^2 2t - 2 \sin 2t \cos 2t + 4\sin ^2 2t )$

$x'^2 + y'^2 = 25e^t (1/4 + 4 )$ and this is easy to integrate

5. this is wrong, the exponents are negative

6. Originally Posted by apples12
this is wrong, the exponents are negative
so? ... take Amer's post and fix the signs.